Final answer:
The heat required to vaporize 84.8 g of water at its boiling point is 191 kJ when using the latent heat of vaporization for water, which is 2260 J/g.
Step-by-step explanation:
To calculate the amount of heat required to vaporize 84.8 g of water at its boiling point, we should use the latent heat of vaporization of water, which is the energy needed for the phase change from liquid to gas at constant temperature.
The latent heat of vaporization of water is 2260 J/g. Therefore, we can find the total heat (Q) needed using the formula:
Q = m×Lv
Where:
- m = mass of the water (84.8 g)
- Lv = latent heat of vaporization of water (2260 J/g)
Plugging in the values, we get:
Q = 84.8 g × 2260 J/g
Q = 191648 J
Converting Joules to kilojoules (since 1 kJ = 1000 J):
Q = 191.648 kJ
To three significant figures, the heat required to vaporize 84.8 g of water at its boiling point is 191 kJ.